Page 379 - Analysis and Design of Machine Elements

Sliding Bearings
Steps Computation Results Units 357

4. Compute Unit During start-up, the unit load will be the static load. p = 1.511 MPa
bearing load From Eq. (12.3)
F 17000
p = = = 1.511 MPa
dB 150 × 75
5. Select bearing Select tin bronze ZCuSn10P1 to ensure p ≤ [p], v ≤ [v], Tin bronze
material pv ≤ [pv] ZCuSn10P1
6. Estimate the From Eq. (12.29)
dynamic viscosity of = 0.068 = 0.068 = 0.0266 Pa ⋅ s
the lubricant n 1∕3 (1000∕60) 1∕3
3
7. Compute the Assuming the density = 900 kg∕m .
kinematic viscosity
From Eq. (12.2)
0.0266
6
6
= × 10 = × 10 = 29.58 cSt
900
∘
8. Select average t = 50 C
m
temperature
9. Select lubricating Select lubricating oil L-AN 46 from Figure 12.5. lubricating
oil oil L-AN 46
10. select the From Figure 12.5, select the kinematic viscosity of 50 = 30 cSt
∘
kinematic viscosity of L-AN46 at t = 50 Cas 50 = 30 cSt.
m
lubricant
∘
11. Compute the Compute the dynamic viscosity of L-AN64 at 50 C = 0.027 Pa s
dynamic viscosity = ×10 −6 = 900 × 30 × 10 −6 =0.027 Pa s
50 50
12. Estimate the From Eq. (12.30) = 0.001339
√
relative clearance = 0.8 v × 10 −3 = 0.8 × √ 7.85 × 10 −3 = 0.001339
4
4
13. Compute the Δ= d =0.001339 × 150 = 0.20 mm Δ = 0.20 mm
diametrical clearance
14. Compute the load From Eq. (12.23) C = 0.9587
p
carrying capacity C = F 2 = 17000 ×(0.001339) 2 = 0.9587
coefficient p 2 vB 2 × 0.027 × 7.85 × 0.075
15. Compute the According to C and B/d,from = 0.676
p
eccentricity ratio
Table 12.2, by interpolating, we have = 0.676
16. Compute the From Eq. (12.24), the minimum film thickness is h min = 32.5 μm
minimum film h = d (1 − )= 150 × 0.001339 ×(1 − 0.676)
thickness min 2 2
= 0.0325 mm
17. Determine the Select R = 0.0032 mm for the journal and R = 0.0032 mm
z1 z1
surface roughness for R = 0.0063 mm for the bearing R = 0.0063 mm
z2
the journal and z2
bearing
18. Compute the Assume a safety factor of 2.0, the allowable film [h] = 19 μm
allowable film thickness is calculated as
thickness [h]= S(R + R )=2 × (0.0032 + 0.0063)
z1 z2
=0.019 mm
Since h min > [h], the design is satisfactory.
(continued)